### Access Answers of Maths NCERT Class 8 Chapter 6- Squares and Square Roots Exercise 6.4 Page Number 107

**1. Find the square root of each of the following numbers by Division method.**

**i. 2304**

**ii. 4489**

**iii. 3481**

**iv. 529**

**v. 3249**

**vi. 1369**

**vii. 5776**

**viii. 7921**

**ix. 576**

**x. 1024**

**xi. 3136**

**xii. 900**

Solution:

i.

∴ √2304 = 48

ii.

∴ √4489 = 67

iii.

∴ √3481 = 59

iv.

∴ √529 = 23

∴ √3249 = 57

vi.

∴ √1369 = 37

∴ √5776 = 76

∴ √7921 = 89

ix.

∴ √576 = 24

x.

∴ √1024 = 32

xi.

∴ √3136 = 56

xii.

∴ √900 = 30

**2. Find the number of digits in the square root of each of the following numbers (without any**

**calculation).64**

**i. 144**

**ii. 4489**

**iii. 27225**

**iv. 390625**

Solution:

i.

∴ √144 = 12

Hence, the square root of the number 144 has 2 digits.

ii.

∴ √4489 = 67

Hence, the square root of the number 4489 has 2 digits.

iii.

√27225 = 165

Hence, the square root of the number 27225 has 3 digits.

∴ √390625 = 625

Hence, the square root of the number 390625 has 3 digits.

**3. Find the square root of the following decimal numbers.**

**i. 2.56**

**ii. 7.29**

**iii. 51.84**

**iv. 42.25**

**v. 31.36**

Solution:

**i.**

∴ √2.56 = 1.6

ii.

∴ √7.29 = 2.7

iii.

∴ √51.84 = 7.2

iv.

∴ √42.25 = 6.5

∴ √31.36 = 5.6

**4. Find the least number which must be subtracted from each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.**

**i. 402**

**ii. 1989**

**iii. 3250**

**iv. 825**

**v. 4000**

Solution:

i.

∴ We must subtract 2 from 402 to get a perfect square.

New number = 402 – 2 = 400

∴ √400 = 20

ii.

∴ We must subtract 53 from 1989 to get a perfect square. New number = 1989 – 53 = 1936

∴ √1936 = 44

iii.

∴ We must subtract 1 from 3250 to get a perfect square.

New number = 3250 – 1 = 3249

∴ √3249 = 57

iv.

We must subtracted 41 from 825 to get a perfect square.

New number = 825 – 41 = 784

∴ √784 = 28

∴ We must subtract 31 from 4000 to get a perfect square. New number = 4000 – 31 = 3969

∴ √3969 = 63

**5. Find the least number which must be added to each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.**

**(i) 525**

**(ii) 1750**

**(iii) 252**

**(iv)1825**

**(v)6412**

Solution:

(i)

Here, (22)2 < 525 > (23)2

We can say 525 is ( 129 – 125 ) 4 less than (23)2.

∴ If we add 4 to 525, it will be perfect square. New number = 525 + 4 = 529

∴ √529 = 23

Here, (41)2 < 1750 > (42)^{2}

We can say 1750 is ( 164 – 150 ) 14 less than (42)^{2}.

∴ If we add 14 to 1750, it will be perfect square.

New number = 1750 + 14 = 1764

∴√1764 = 42

(iii)

Here, (15)2 < 252 > (16)2

We can say 252 is ( 156 – 152 ) 4 less than (16)2.

∴ If we add 4 to 252, it will be perfect square.

New number = 252 + 4 = 256

∴ √256 = 16

(iv)

Here, (42)2 < 1825 > (43)2

We can say 1825 is ( 249 – 225 ) 24 less than (43)2.

∴ If we add 24 to 1825, it will be perfect square.

New number = 1825 + 24 = 1849

∴ √1849 = 43

(v)

Here, (80)2 < 6412 > (81)2

We can say 6412 is ( 161 – 12 ) 149 less than (81)2.

∴ If we add 149 to 6412, it will be perfect square.

New number = 6412 + 149 = 656

∴ √6561 = 81

**6. Find the length of the side of a square whose area is 441 m2.**

Solution:

Let the length of each side of the field = a Then, area of the field = 441 m2

⇒ a2 = 441 m2

⇒a = √441 m

∴ The length of each side of the field = a m = 21 m.

**7. In a right triangle ABC, ∠B = 90°.**

**a. If AB = 6 cm, BC = 8 cm, find AC**

**b. If AC = 13 cm, BC = 5 cm, find AB**

Solution:

a.

Given, AB = 6 cm, BC = 8 cm

Let AC be x cm.

∴ AC2 = AB2 + BC2

Hence, AC = 10 cm.

b.

Given, AC = 13 cm, BC = 5 cm

Let AB be x cm.

∴ AC2 = AB2 + BC2

⇒ AC2 – BC2 = AB2

Hence, AB = 12 cm

**8. A gardener has 1000 plants. He wants to plant these in such a way that the number of rows**

**and the number of columns remain same. Find the minimum number of plants he needs more for this.**

Solution:

Let the number of rows and column be, x.

∴ Total number of row and column= x× x = x2 As per question, x2 = 1000

⇒ x = √1000

Here, (31)2 < 1000 > (32)2

We can say 1000 is ( 124 – 100 ) 24 less than (32)2.

∴ 24 more plants are needed.

**9. There are 500 children in a school. For a P.T. drill they have to stand in such a manner that the number of rows is equal to number of columns. How many children would be left out in this arrangement.**

Solution:

Let the number of rows and column be, x.

∴ Total number of row and column= x × x = x2 As per question, x2 = 500

x = √500

Hence, 16 children would be left out in the arrangement